*This preview shows
pages
1–4. Sign up to
view the full content.*

1
SOLUTIONS TO PROBLEMS
Easy
P19.2
(a)
N
=
10.0 grams
107.87 grams mol
!
"
#
$
%
&
6.02
'
10
23
atoms
mol
!
"
#
$
%
&
47
electrons
atom
!
"
#
$
%
&
=
2.62
'
10
24
(b)
# electrons added
=
Q
e
=
1.00
!
10
"
3
C
1.60
!
10
"
19
C electron
=
6.25
!
10
15
or
2.38 electrons for every 10
9
already present
.
P19.4
The force on one proton is
r
F
=
k
e
q
1
q
2
r
2
away from the other proton. Its magnitude is
8.99
!
10
9
N
"
m C
2
( )
1.6
!
10
#
19
C
2
!
10
#
15
m
$
%
&
'
(
)
2
=
57.5 N
.
P19.7
(a)
The force is one of
attraction
. The distance
r
in Coulomb’s law is the distance between
centers. The magnitude of the force is
F
=
k
e
q
1
q
2
r
2
=
8.99
!
10
9
N
"
m
2
C
2
( )
12.0
!
10
#
9
C
( )
18.0
!
10
#
9
C
( )
0.300 m
( )
2
=
2.16
!
10
#
5
N
.
(b)
The net charge of
!
6.00
"
10
!
9
C
will be equally split between the two spheres, or
!
3.00
"
10
!
9
C
on each. The force is one of
repulsion
, and its magnitude is
F
=
k
e
q
1
q
2
r
2
=
8.99
!
10
9
N
"
m
2
C
2
( )
3.00
!
10
#
9
C
( )
3.00
!
10
#
9
C
( )
0.300 m
( )
2
=
8.99
!
10
#
7
N
.

This ** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*2
Electric Forces and Electric Fields
P19.10
For equilibrium,
r
F
e
=
!
r
F
g
or
q
r
E
=
!
mg
!
ˆ
j
( )
.
Thus,
r
E
=
mg
q
ˆ
j
.
(a)
r
E
=
mg
q
ˆ
j
=
9.11
!
10
"
31
kg
( )
2
( )
"
1.60
!
10
"
19
C
( )
ˆ
j
=
"
5.58
!
10
"
11
( )
ˆ
j
(b)
r
E
=
mg
q
ˆ
j
=
1.67
!
10
"
27
kg
( )
2
( )
1.60
!
10
"
19
C
( )
ˆ
j
=
1.02
!
10
"
7
( )
ˆ
j
*P19.24
(a)
q
1
q
2
=
!
6
18
=
!
1
3
(b)
q
1
is negative,
q
2
is positive
P19.27
(a)
a
=
qE
m
=
1.60
!
10
"
19
640
( )
1.67
!
10
"
27
=
6.13
!
10
10
2
(b)
v
f
=
v
i
+
at
1.20
!
10
6
=
6.13
!
10
10
( )
t
t
=
1.95
!
10
"
5
s
(c)
x
f
!
x
i
=
1
2
v
i
+
v
f
( )
t
x
f
=
1
2
1.20
!
10
6
( )
1.95
!
10
"
5
( )
=
11.7 m
(d)
K
=
1
2
mv
2
=
1
2
1.67
!
10
"
27
kg
( )
1.20
!
10
6
( )
2
=
1.20
!
10
"
15
J
P19.30
!
E
=
EA
cos
"
=
2.00
#
10
4
( )
18.0 m
2
( )
cos10.0
° =
355 kN
$
m
2
P19.31
!
E
=
EA
cos
A
=
!
r
2
=
0.200
( )
2
=
0.126 m
2
5.20
!
10
5
=
E
0.126
( )
cos0
°
E
=
4.14
!
10
6
N C
=
4.14 MN C
P19.32
(a)
E
=
k
e
Q
r
2
:
8.90
!
10
2
=
8.99
!
10
9
( )
Q
0.750
( )
2
But
Q
is negative since
r
E
points inward.
Q
=
!
5.57
"
10
!
8
C
=
!
55.7 nC
(b)
The
negative
charge has a
spherically symmetric
charge distribution, concentric with
the spherical shell.